3.2.3 \(\int \frac {(1-x^3)^{2/3}}{(1+x+x^2)^2} \, dx\) [103]

Optimal. Leaf size=43 \[ \frac {1}{\sqrt [3]{1-x^3}}+\frac {x}{\sqrt [3]{1-x^3}}-x^2 \, _2F_1\left (\frac {2}{3},\frac {4}{3};\frac {5}{3};x^3\right ) \]

[Out]

1/(-x^3+1)^(1/3)+x/(-x^3+1)^(1/3)-x^2*hypergeom([2/3, 4/3],[5/3],x^3)

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Rubi [A]
time = 0.03, antiderivative size = 43, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {2183, 197, 371, 267} \begin {gather*} x^2 \left (-\, _2F_1\left (\frac {2}{3},\frac {4}{3};\frac {5}{3};x^3\right )\right )+\frac {x}{\sqrt [3]{1-x^3}}+\frac {1}{\sqrt [3]{1-x^3}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(1 - x^3)^(2/3)/(1 + x + x^2)^2,x]

[Out]

(1 - x^3)^(-1/3) + x/(1 - x^3)^(1/3) - x^2*Hypergeometric2F1[2/3, 4/3, 5/3, x^3]

Rule 197

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[x*((a + b*x^n)^(p + 1)/a), x] /; FreeQ[{a, b, n, p}, x] &
& EqQ[1/n + p + 1, 0]

Rule 267

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(a + b*x^n)^(p + 1)/(b*n*(p + 1)), x] /; FreeQ
[{a, b, m, n, p}, x] && EqQ[m, n - 1] && NeQ[p, -1]

Rule 371

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[a^p*((c*x)^(m + 1)/(c*(m + 1)))*Hyperg
eometric2F1[-p, (m + 1)/n, (m + 1)/n + 1, (-b)*(x^n/a)], x] /; FreeQ[{a, b, c, m, n, p}, x] &&  !IGtQ[p, 0] &&
(ILtQ[p, 0] || GtQ[a, 0])

Rule 2183

Int[(Px_.)*((c_) + (d_.)*(x_) + (e_.)*(x_)^2)^(q_)*((a_) + (b_.)*(x_)^3)^(p_.), x_Symbol] :> Dist[1/c^q, Int[E
xpandIntegrand[(c^3 - d^3*x^3)^q*(a + b*x^3)^p, Px/(c - d*x)^q, x], x], x] /; FreeQ[{a, b, c, d, e, p}, x] &&
PolyQ[Px, x] && EqQ[d^2 - c*e, 0] && ILtQ[q, 0] && RationalQ[p] && EqQ[Denominator[p], 3]

Rubi steps

\begin {align*} \int \frac {\left (1-x^3\right )^{2/3}}{\left (1+x+x^2\right )^2} \, dx &=\int \left (-\frac {4 \left (1-x^3\right )^{2/3}}{3 \left (-1+i \sqrt {3}-2 x\right )^2}+\frac {4 i \left (1-x^3\right )^{2/3}}{3 \sqrt {3} \left (-1+i \sqrt {3}-2 x\right )}-\frac {4 \left (1-x^3\right )^{2/3}}{3 \left (1+i \sqrt {3}+2 x\right )^2}+\frac {4 i \left (1-x^3\right )^{2/3}}{3 \sqrt {3} \left (1+i \sqrt {3}+2 x\right )}\right ) \, dx\\ &=-\left (\frac {4}{3} \int \frac {\left (1-x^3\right )^{2/3}}{\left (-1+i \sqrt {3}-2 x\right )^2} \, dx\right )-\frac {4}{3} \int \frac {\left (1-x^3\right )^{2/3}}{\left (1+i \sqrt {3}+2 x\right )^2} \, dx+\frac {(4 i) \int \frac {\left (1-x^3\right )^{2/3}}{-1+i \sqrt {3}-2 x} \, dx}{3 \sqrt {3}}+\frac {(4 i) \int \frac {\left (1-x^3\right )^{2/3}}{1+i \sqrt {3}+2 x} \, dx}{3 \sqrt {3}}\\ \end {align*}

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Mathematica [A]
time = 10.11, size = 43, normalized size = 1.00 \begin {gather*} \frac {(1+2 x) \left (1-x^3\right )^{2/3}}{1+x+x^2}+x^2 \, _2F_1\left (\frac {1}{3},\frac {2}{3};\frac {5}{3};x^3\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(1 - x^3)^(2/3)/(1 + x + x^2)^2,x]

[Out]

((1 + 2*x)*(1 - x^3)^(2/3))/(1 + x + x^2) + x^2*Hypergeometric2F1[1/3, 2/3, 5/3, x^3]

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Mathics [F(-1)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Warning: Unable to verify antiderivative.

[In]

mathics('Integrate[(1 - x^3)^(2/3)/(1 + x + x^2)^2,x]')

[Out]

Timed out

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Maple [A]
time = 0.21, size = 34, normalized size = 0.79

method	result	size
risch	\(-\frac {\left (-1+x \right ) \left (1+2 x \right )}{\left (-x^{3}+1\right )^{\frac {1}{3}}}+x^{2} \hypergeom \left (\left [\frac {1}{3}, \frac {2}{3}\right ], \left [\frac {5}{3}\right ], x^{3}\right )\)	\(34\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-x^3+1)^(2/3)/(x^2+x+1)^2,x,method=_RETURNVERBOSE)

[Out]

-(-1+x)*(1+2*x)/(-x^3+1)^(1/3)+x^2*hypergeom([1/3,2/3],[5/3],x^3)

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-x^3+1)^(2/3)/(x^2+x+1)^2,x, algorithm="maxima")

[Out]

integrate((-x^3 + 1)^(2/3)/(x^2 + x + 1)^2, x)

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Fricas [F]
time = 0.30, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-x^3+1)^(2/3)/(x^2+x+1)^2,x, algorithm="fricas")

[Out]

integral((-x^3 + 1)^(2/3)/(x^4 + 2*x^3 + 3*x^2 + 2*x + 1), x)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (- \left (x - 1\right ) \left (x^{2} + x + 1\right )\right )^{\frac {2}{3}}}{\left (x^{2} + x + 1\right )^{2}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-x**3+1)**(2/3)/(x**2+x+1)**2,x)

[Out]

Integral((-(x - 1)*(x**2 + x + 1))**(2/3)/(x**2 + x + 1)**2, x)

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Giac [F] N/A
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-x^3+1)^(2/3)/(x^2+x+1)^2,x)

[Out]

Could not integrate

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.02 \begin {gather*} \int \frac {{\left (1-x^3\right )}^{2/3}}{{\left (x^2+x+1\right )}^2} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((1 - x^3)^(2/3)/(x + x^2 + 1)^2,x)

[Out]

int((1 - x^3)^(2/3)/(x + x^2 + 1)^2, x)

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